Mathematics 9709 · Probability & Statistics 1

Standard deviation

Standard deviation is the typical distance of a value from the mean.
We square the gaps so values above and below the mean do not cancel.
Then we average the squared gaps and square-root back to the original units.
The exam shortcut is: mean of the squares minus the square of the mean.

The variance machine: totals in, spread out

$\operatorname{Var}=\dfrac{\sum x^2}{n}-\bar x^2,\qquad \text{SD}=\sqrt{\operatorname{Var}},\qquad \bar x=\dfrac{\sum x}{n}$

For a frequency table, build the columns $x\mid f\mid xf\mid x^2f$ and use $\operatorname{Var}=\dfrac{\sum x^2 f}{\sum f}-\bar x^2$ . For grouped data, $x$ is the class mid-value and the answer is an estimate.

Var = mean of the squares − square of the mean. Three numbers run it: n, Σx, Σx².

Columns $x\mid f\mid xf\mid x^2f$ (compute $x^2f$ as $x^2\cdot f$ , never $(xf)^2$ ).
Totals $\sum f,\ \sum xf,\ \sum x^2f$ .
$\bar x=\sum xf/\sum f$ — keep it exact.
$\operatorname{Var}=\sum x^2f/\sum f-\bar x^2$ ; $\text{SD}=\sqrt{\operatorname{Var}}$ (don't forget the root).

Worked example

SD of a raw list

2, 4, 4, 6, 8, 10: $n=6$ , $\sum x=34$ , $\sum x^2=236$ .

$\operatorname{Var}=\tfrac{236}{6}-\big(\tfrac{34}{6}\big)^2=39.33-32.11=7.22$

Answer

\text{SD}=\sqrt{7.22}=2.69

Worked example

SD + which values lie more than 1 SD from the mean

3, 9, 15, 24, 29: $n=5$ , $\sum x=80$ , $\sum x^2=1732$ , so $\bar x=16$ .

$\operatorname{Var}=\tfrac{1732}{5}-16^2=346.4-256=90.4,\quad \text{SD}=\sqrt{90.4}=9.51$

“More than 1 SD from the mean” means outside $16\pm9.51$ , i.e. outside $(6.49,\ 25.51)$ .

Answer

\text{SD}=9.51;\quad 3\text{ and }29\text{ lie outside }16\pm9.51

“Within 1 SD” = the band x̄ ± SD; test each value against it. A very common 1-mark follow-up.

Worked example

SD from a discrete frequency table

$x=12,14,16$ with $f=13,28,10$ . Columns give $\sum f=51,\ \sum xf=708,\ \sum x^2f=9920$ (build $x^2f$ as $x\cdot xf$ : e.g. $12\times156=1872$ ).

$\operatorname{Var}=\tfrac{9920}{51}-\big(\tfrac{708}{51}\big)^2=194.5-192.7=1.79$

Answer

\text{SD}=\sqrt{1.79}=1.34

Worked example

SD from a grouped frequency table

Mid-values $x=5,15,25,35$ with $f=6,10,9,5$ : $\sum f=30,\ \sum xf=580,\ \sum x^2f=14150$ , so $\bar x=\tfrac{580}{30}=\tfrac{58}{3}$ .

$\operatorname{Var}=\tfrac{14150}{30}-\big(\tfrac{58}{3}\big)^2=471.67-373.78=97.89$

Answer

\text{SD}=\sqrt{97.89}\approx 9.89\ (\text{an estimate})

Want	Pick	Note
Uses all data, algebra-friendly, combinable	SD	but one outlier inflates it
Resistant to extremes	IQR	for skewed data / box plots

Examiner note

The method mark is the full variance formula with values substituted (

\sum x^2f/\sum f-\bar x^2

shown before the root); the final SD must be identified (“SD = …”). The same engine drives a discrete-RV variance,

\operatorname{Var}(X)=\sum x^2p-\big(\sum xp\big)^2

— probabilities in place of frequencies. (9709/52/M/J/24 Q5)

Common mistake

Compute

x^2f

x^2\cdot f

, never

(xf)^2

;

\sum x^2

is “square then add”, not

(\sum x)^2

. Giving the variance as the final answer when the SD was asked (no √).

Why carry the exact mean

The mean enters the variance squared, so a small rounding gets amplified. In the table above the exact mean gives SD 1.34, but rounding $\bar x$ to 13.9 first gives 1.14 — out by about 15%. Keep $\bar x$ as a fraction (or full calculator value) until the very end.

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Standard deviation

Standard deviation is the typical distance of a value from the mean.
We square the gaps so values above and below the mean do not cancel.
Then we average the squared gaps and square-root back to the original units.
The exam shortcut is: mean of the squares minus the square of the mean.

The variance machine: totals in, spread out

$\operatorname{Var}=\dfrac{\sum x^2}{n}-\bar x^2,\qquad \text{SD}=\sqrt{\operatorname{Var}},\qquad \bar x=\dfrac{\sum x}{n}$

Var = mean of the squares − square of the mean. Three numbers run it: n, Σx, Σx².

Columns $x\mid f\mid xf\mid x^2f$ (compute $x^2f$ as $x^2\cdot f$ , never $(xf)^2$ ).
Totals $\sum f,\ \sum xf,\ \sum x^2f$ .
$\bar x=\sum xf/\sum f$ — keep it exact.
$\operatorname{Var}=\sum x^2f/\sum f-\bar x^2$ ; $\text{SD}=\sqrt{\operatorname{Var}}$ (don't forget the root).

Worked example

SD of a raw list

2, 4, 4, 6, 8, 10: $n=6$ , $\sum x=34$ , $\sum x^2=236$ .

$\operatorname{Var}=\tfrac{236}{6}-\big(\tfrac{34}{6}\big)^2=39.33-32.11=7.22$

Answer

\text{SD}=\sqrt{7.22}=2.69

Worked example

SD + which values lie more than 1 SD from the mean

3, 9, 15, 24, 29: $n=5$ , $\sum x=80$ , $\sum x^2=1732$ , so $\bar x=16$ .

$\operatorname{Var}=\tfrac{1732}{5}-16^2=346.4-256=90.4,\quad \text{SD}=\sqrt{90.4}=9.51$

“More than 1 SD from the mean” means outside $16\pm9.51$ , i.e. outside $(6.49,\ 25.51)$ .

Answer

\text{SD}=9.51;\quad 3\text{ and }29\text{ lie outside }16\pm9.51

“Within 1 SD” = the band x̄ ± SD; test each value against it. A very common 1-mark follow-up.

Worked example

SD from a discrete frequency table

$x=12,14,16$ with $f=13,28,10$ . Columns give $\sum f=51,\ \sum xf=708,\ \sum x^2f=9920$ (build $x^2f$ as $x\cdot xf$ : e.g. $12\times156=1872$ ).

$\operatorname{Var}=\tfrac{9920}{51}-\big(\tfrac{708}{51}\big)^2=194.5-192.7=1.79$

Answer

\text{SD}=\sqrt{1.79}=1.34

Worked example

SD from a grouped frequency table

Mid-values $x=5,15,25,35$ with $f=6,10,9,5$ : $\sum f=30,\ \sum xf=580,\ \sum x^2f=14150$ , so $\bar x=\tfrac{580}{30}=\tfrac{58}{3}$ .

$\operatorname{Var}=\tfrac{14150}{30}-\big(\tfrac{58}{3}\big)^2=471.67-373.78=97.89$

Answer

\text{SD}=\sqrt{97.89}\approx 9.89\ (\text{an estimate})

Want	Pick	Note
Uses all data, algebra-friendly, combinable	SD	but one outlier inflates it
Resistant to extremes	IQR	for skewed data / box plots

Examiner note

The method mark is the full variance formula with values substituted (

\sum x^2f/\sum f-\bar x^2

shown before the root); the final SD must be identified (“SD = …”). The same engine drives a discrete-RV variance,

\operatorname{Var}(X)=\sum x^2p-\big(\sum xp\big)^2

— probabilities in place of frequencies. (9709/52/M/J/24 Q5)

Common mistake

Compute

x^2f

x^2\cdot f

, never

(xf)^2

;

\sum x^2

is “square then add”, not

(\sum x)^2

. Giving the variance as the final answer when the SD was asked (no √).

Why carry the exact mean